Problem 9
Question
Let's use the symbol \(*\) for a new type of multiplication of complex numbers defined by \(z_{1} * z_{2}=\left(x_{1} x_{2}, y_{1} y_{2}\right) .\) This exercise shows why this is an unfortunate definition. (a) Use the definition given in property (P7) and state what the multiplicative identity \(\zeta\) would have to be for this new multiplication. (b) Show that if you use this new multiplication, nonzero complex numbers of the form \((0, a)\) have no inverse. That is, show that if \(z=(0, a)\), there is no complex number \(w\) with the property that \(z * w=\zeta\), where C is the multiplicative identity you found in part (a).
Step-by-Step Solution
Verified Answer
(a) The identity is (1, 1). (b) No inverse exists for (0, a).
1Step 1: Determine the Multiplicative Identity
For a complex number multiplication to have a multiplicative identity, we need a number \(\zeta = (x, y)\) such that for any complex number \(z_1 = (a, b)\), the operation \(z_1 * \zeta = (a, b)\). Using the given operation, this implies \( (a \cdot x, b \cdot y) = (a, b) \). For this to hold true for all \(a\) and \(b\), we must have \(x = 1\) and \(y = 1\). Hence, the multiplicative identity \(\zeta\) is \( (1, 1) \).
2Step 2: Prove Nonexistence of Inverse for (0, a)
We need to show there is no complex number \(w = (c, d)\) such that \((0, a) * w = \zeta = (1, 1)\). Applying the new multiplication, \((0 \cdot c, a \cdot d) = (0, a)\), we have \(0 = 1\) and \(a \cdot d = 1\). The first condition, \(0 = 1\), is false no matter the choice of \(c\) or \(d\), leading to a contradiction. Therefore, no such \(w\) exists.
Key Concepts
Multiplicative IdentityInverse of Complex NumbersNon standard multiplication rules
Multiplicative Identity
In mathematics, the concept of a multiplicative identity is crucial as it represents a number that, when multiplied by any other number, does not change the value of that number. In the realm of standard complex numbers, the multiplicative identity is typically represented as 1. However, when exploring custom operations like in this exercise, we must redefine this identity according to the rules given.
In the problem at hand, the new multiplication is defined by combining the real parts and imaginary parts of the complex numbers separately: \(z_1 * z_2 = (x_1 x_2, y_1 y_2)\). To determine the identity element \(\zeta\), we need \(z_1 * \zeta = z_1\) for every complex number \(z_1 = (a, b)\). This requires that the identity must result in \((a \cdot x, b \cdot y) = (a, b)\).
From this, we find that \(x\) and \(y\) must both be 1. Therefore, for this non-standard multiplication operation, the multiplicative identity is \((1, 1)\). It essentially means that both components must independently preserve themselves when multiplied by the identity.
In the problem at hand, the new multiplication is defined by combining the real parts and imaginary parts of the complex numbers separately: \(z_1 * z_2 = (x_1 x_2, y_1 y_2)\). To determine the identity element \(\zeta\), we need \(z_1 * \zeta = z_1\) for every complex number \(z_1 = (a, b)\). This requires that the identity must result in \((a \cdot x, b \cdot y) = (a, b)\).
From this, we find that \(x\) and \(y\) must both be 1. Therefore, for this non-standard multiplication operation, the multiplicative identity is \((1, 1)\). It essentially means that both components must independently preserve themselves when multiplied by the identity.
Inverse of Complex Numbers
The inverse of a number is a value that, when multiplied by the original number, yields the multiplicative identity. For complex numbers in the traditional sense, finding an inverse involves using both real and imaginary parts to "undo" each other. However, in our alternate multiplication setup, things become a bit tricky.
In the exercise, we're asked to examine numbers of the form \((0, a)\). If we seek an inverse \(w = (c, d)\) for \((0, a)\) so that \((0, a) * w = \zeta = (1, 1)\), we run into immediate issues. The multiplication gives us \((0 \cdot c, a \cdot d)\). To satisfy the identity \( (1, 1)\), we require \(0 = 1\) and \(a \cdot d = 1\).
In the exercise, we're asked to examine numbers of the form \((0, a)\). If we seek an inverse \(w = (c, d)\) for \((0, a)\) so that \((0, a) * w = \zeta = (1, 1)\), we run into immediate issues. The multiplication gives us \((0 \cdot c, a \cdot d)\). To satisfy the identity \( (1, 1)\), we require \(0 = 1\) and \(a \cdot d = 1\).
- The equation \(0 = 1\) is clearly false, creating an unsolvable contradiction.
- This shows that the inverse for \((0, a)\) cannot exist under this multiplication.
Non standard multiplication rules
In mathematics, rules play a critical role in determining how numbers interact with one another. The traditional multiplication of complex numbers involves distributing and combining both their real and imaginary components using the rule \((a + bi)(c + di) = (ac - bd) + (ad + bc)i\). However, this exercise explores an atypical method where multiplication is redefined.
Under this new operation, \(z_1 * z_2 = (x_1 x_2, y_1 y_2)\), each component is multiplied individually rather than intermixing real and imaginary parts. While this seems straightforward, it introduces several complications, especially with established properties like identity and inverses.
Under this new operation, \(z_1 * z_2 = (x_1 x_2, y_1 y_2)\), each component is multiplied individually rather than intermixing real and imaginary parts. While this seems straightforward, it introduces several complications, especially with established properties like identity and inverses.
- For instance, multiplicative identity must satisfy both component multiplications individually, leading to \((1, 1)\).
- Moreover, many complex numbers like \((0, a)\), which work fine under standard rules, suddenly lose their ability to have inverses.
Other exercises in this chapter
Problem 8
Show that the equation \(\operatorname{Arg}\left(z_{1} z_{2}\right)=\) Arg \(z_{1}+\) Arg \(z_{2}\) is true if \(\frac{-\pi}{2}
View solution Problem 8
Solve the equation \((z+1)^{3}=z^{3}\).
View solution Problem 9
Show that \(\left|z_{1}-z_{2}\right| \leq\left|z_{1}\right|+\left|z_{2}\right|\)
View solution Problem 9
Describe the set of complex numbers for which \(\operatorname{Arg}\left(\frac{1}{x}\right) \neq-\operatorname{Arg}(z) .\) Prove your assertion.
View solution